Multiple Choice QuestionsA pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is
more than 3 but less than 6
more than 6 but less than 9
more than 9
more than 9
A.
more than 3 but less than 6
To reverse the direction
A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is
g
2/3g
g/3
g/3
B.
2/3g
A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as
17.6 kg ms−1
17.565 kg ms−1
17.56 kg ms−1
17.56 kg ms−1
A.
17.6 kg ms−1
P = mv = 3.513 × 5.00 ≈ 17.6
For a particle in uniform circular motion, the acceleration an at 
point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)
C.
Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?
4.9 ms-2 in horizontal direction
9.8 ms-2 in vertical direction
zero
zero
D.
zero
For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of A with respect to B is
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)
A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is
A.
According to Newton's law of cooling.
Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.
A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
0.16 J
1.00 J
0.67 J
0.67 J
C.
0.67 J
m1u1 + m2u2 = (m1 + m2)v
v = 2/3 m/s
STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.
The statement I is True, Statement II is False.
The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.
The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.
The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.
B.
The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.
A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
K
0
K/2
K/2
D.
K/2
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is
0.0125 Nm-1
0.1 Nm-1
0.05 Nm-1
0.05 Nm-1
D.
0.05 Nm-1
⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m
